OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Ex 16(g)

Students can track their progress and improvement through regular use of Class 11 ISC Maths S Chand Solutions Chapter 16 The Straight Line Ex 16(g).

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(g)

Question 1.
Find the equations of the lines bisecting the angles between the following pairs of straight lines writing first at the bisector of the angle in which the origin lines :
(i) 3x – 4y + 10 = 0, 5x – 12y – 10 = 0;
(ii) 12x – 5y + 3 = 0, 4x + 3y – 2 = 0.
Solution:
(i) Writing given lines with positive constant terms, we get
3x – 4y + 10 = 0 …(1)
-5x + 12y + 10 = 0 …(2)
The equations of the angular bisectors of lines (1) and (2) are given by

Thus the eqn. of bisector in which origin lies (Taking positive sign) is given by
\(\frac{3 x-4 y+10}{5}\) = \(\frac{-5 x+12 y+10}{13}\)
⇒ 39x – 52y + 130 = – 25x + 60y + 50
⇒ 64x – 112y + 80 = 0
⇒ 4x – 7y + 5 = 0
The eqn. of other bisector be given by
\(\frac{3 x-4 y+10}{5}\) = \(-\frac{(-5 x+12 y+10)}{13}\)
⇒ 39x – 52y + 130 = 25x – 60y – 50
⇒ 14x + 8y + 180 = 0
⇒ 7x + 4y + 90 = 0

(ii) Writing the given eqns. with positive constant terms we have
12x – 5y + 3 = 0 …(1)
-4x – 3y + 2 = 0 …(2)
The equations of angular bisectors of lines (1) and (2) be given by

The bisector of the angle in which origin lies is given by
\(\frac{12 x-5 y+3}{13}\) = \(\frac{-4 x-3 y+2}{5}\)
⇒ 60x – 25y + 15 = – 52x – 39y + 26
⇒ 112x + 14y – 11 = 0
and the eqn. of other bisector be given by
\(\frac{12 x-5 y+3}{13}\) = –\(\left(\frac{-4 x-3 y+2}{5}\right)\)
⇒ 60x – 25y + 15 = + 52x + 39y – 26
⇒ 8x – 64y + 41 = 0

Question 2.
Find the equations of the bisectors of the angles between 4x + 3y – 4 = 0 and 12x + 5y – 3 = 0. Show that these bisectors are at right angles to each other.
Solution:
Eqns. of given lines are
4x + 3y – 4 = 0 …(1)
12x + 5y – 3 = 0 …(2)
The eqn’s of angular bisector of lines (1) and (2) be given by

Taking +ve sign ;
\(\frac{4 x+3 y-4}{5}\) = \(\frac{12 x+5 y-3}{13}\)
⇒ 52x + 39y – 52 = 60x + 25y – 15
⇒ 8x – 14y + 37 = 0 …(3)
Taking -ve sign ;
\(\frac{4 x+3 y-4}{5}\) = –\(\left(\frac{12 x+5 y-3}{13}\right)\)
⇒ 52x + 39x – 52 = – 60x – 25y + 15
⇒ 112x + 64x – 67 = 0 …(1)
slope of line (3) = m₁ = \(\frac{-8}{-14}\) = \(\frac{4}{7}\)
and slope of bisector (4) = m₂ = –\(\frac{112}{64}\) = –\(\frac{7}{4}\)
Here m₁m₂ = \(\frac{4}{7}\)\(\left(-\frac{7}{4}\right)\) = -1
Thus both bisector are at right angle to each other.

Question 3.
Find the locus of a point which moves so that the perpendiculars drawn from it to the two straight lines 3x + 4y = 5, 12x – 5y = 13 are equal.
Solution:
Given lines are
3x + 4y – 5 = 0 …(1)
and 12x – 5y – 13 = 0 …(2)
Let P(x, y) be any point whose locus is to be find out.
according to given condition we have ⊥ distance from P(x, y) to line (1)
= ⊥ distance from P(x, y) to line (2)

Taking +ve sign ;
13(3x + 4y – 5) = 5(12x – 5y – 13)
⇒ 21x – 77y = 0 ⇒ 3x – 11y = 0
Taking -ve sign ;
13(3x + 4y – 5) = – 5(12x – 5y – 13)
⇒ 99x + 27y – 130 = 0

Question 4.
Find the equations of the lines bisecting the angles between the lines 4x – 3y + 12 = 0 and 12x + 5y = 20. Find, without using the tables, the tangent of an angle between one of these bisectors and one of the origin lines.
Solution:
Eqns. of given lines are
4x – 3y + 12 = 0 …(1)
12x + 5y – 20 = 0 …(2)
The eqns. of angular bisectors of lines (1) and (2) are given by

Taking +ve sign ;
13 (4x – 3y + 12) = 5(12x + 5y – 20)
⇒ 8x + 64y – 256 = 0
⇒ x + 8y – 32 = 0 …(3)
Taking -ve sign ;
13(4x – 3y + 12) = – 5(12x + 5y – 20)
⇒ 112x – 14y + 56 = 0
⇒ 8x – y + 4 = 0 …(4)
Let us take the bisector (3) and line (1)
slope of bisector (3) = –\(\frac { 1 }{ 8 }\) = m₁
slope of line (1) = m₂ = \(\frac{-4}{-3}\) = \(\frac{4}{3}\)
Let θ be the angle between then

Let us take the bisector (4) and line (2).
∴ slope of bisector (4) = m₃ = \(\frac{-8}{-1}\) = 8
and slope of line (2) = m₄ =-\(\frac{12}{5}\)

Question 5.
Find the bisector of the acute angle between the lines :
(i) 3x + 4y = 11 and 12x – 5y = 2;
(ii) 5x = 12y + 24 and 12x = 5y + 10.
Solution:
(i) Writing the given equations with positive constant terms, we have
– 3x – 4y + 11 = 0 …(1)
– 12x + 5y + 2 = 0 …(2)
The eqns. of the angular bisector of lines (1) and (2) be given by

Taking +ve sign :
\(\frac{-3 x-4 y+11}{5}\) = \(\frac{-12 x+5 y+2}{13}\)
⇒ -39x – 52y + 143 = – 60x + 25y + 10
⇒ 21x – 77x + 133 = 0
⇒ 3x – 11x + 19 = 0 …(3)
Taking -ve sign ;
\(\frac{-3 x-4 y+11}{5}\) = \(-\left(\frac{-12 x+5 y+2}{13}\right)\)
⇒ – 39x – 52y + 143 = 60x – 25y – 10
⇒ 99x + 27y – 153 = 0
⇒ 11x + 3y – 17 = 0 …(3)
Out of (3) and (4), that bisector will be the bisector of acute angle which makes an angle θ < 45° with one of lines (1) and (2).
Consider the line (1) and bisector (4)
slope of line = m₁ = –\(\frac{3}{4}\)
slope of bisector (4) = m₂ = –\(\frac{11}{3}\)

Thus the bisector (4) will be bisector of the acute angle between lines (1) and (2).

(ii) Writing the given eqns. with positive constant terms
we have, – 5x + 12y + 24 = 0 …(1)
and – 12x + 5y + 10 = 0 …(2)
The eqns. of angular bisectors of lines (1) and (2) are given by

Taking +ve sign ;
– 5x + 12y + 24 = – 12x + 5y + 10
⇒ 7x + 7y + 14 = 0
⇒ x + y + 2 = 0 …(3)
Taking – ve sign ;
– 5x + 12y + 24 = + 12x – 5y – 10
⇒ 17x – 17y – 34 = 0
⇒ x – y – 2 = 0 …(3)
Out of bisector (3) and (4) that bisector will the bisector of the acute angle which makes an angle θ < 45° with any one of the lines (1) and (2).
Let us take the line (1) and bisector (4).
m₁ = slope of line (1) = –\(\frac{(-5)}{12}\) = \(\frac{5}{12}\)
m₂ = slope of bisector (4) = \(\frac{-1}{-1}\) = 1
If θ be the angle between line (1) and bisector (4)

Thus line (4) gives the bisector of acute angle between lines (1) and (2).

Question 6.
Prove that the perpendiculars drawn from any point of the line 2x + 11y = 5 to the lines 24x + 7y = 20 and 4x – 3y = 2 are equal in length.
Solution:
Equations of given lines are
2x + 11y – 5 = 0 …(1)
24x + 7y – 20 = 0 …(2)
4x – 3y – 2 = 0 …(3)
For any point on line (1) ; we put x = 0 in eqn. (1) ; we have
11y = 5 ⇒ y = \(\frac{5}{11}\)
∴ any point on line (1) be \(\left(0, \frac{5}{11}\right)\).
length of ⊥ from \(\left(0, \frac{5}{11}\right)\) to line (2)

Question 7.
A triangle is formed by the lines whose equations are
AB : x + y – 5 = 0,
BC : x + 7y – 7 = 0,
and CA : 7x + y + 14 = 0 .
Find (i) the bisector of the interior angle at B, and (ii) the bisector of the exterior angle at C.
Solution:
The bisector of the interior angle at B be given by

Taking +ve sign ;
\(\frac{x+7 y-7}{5}\) = x + y – 5
⇒ 4x – 2y – 18 = 0
⇒ 2x – y- 9 = 0 …(3)
Taking -ve sign ;
\(\frac{x+7 y-7}{5}\) = \(-\left(\frac{x+y-5}{1}\right)\)
⇒ x + 7y – 7 = – 5x – 5y + 25
⇒ 6x + 12y – 32 = 0
⇒ 3x + 6y – 16 = 0 …(4)
On solving AB and AC simultaneously; we have
x = –\(\frac { 19 }{ 6 }\) and y = \(\frac { 49 }{ 6 }\)
∴ Coordinates of A are \(\left(-\frac{19}{6}, \frac{49}{6}\right) \text {. }\)
On solving the eqns. of BC and AC; we have
x = \(\frac{-105}{48}\) = \(\frac{-35}{16}\) and y = \(\frac{147}{16 \times 7}\) = \(\frac{21}{16}\)
Out of lines (3) and (4), the line will be the required internal bisector of angle B for which points A and C are lies on its opposite sides.
Substituting the coordinates of points B and C in L.H.S of eqn. (4).

Thus points A and C are lies on opposite side of line (4).
∴ eqn. (4) will be the required internal bisector of angle B.
The bisector of interior angle at B are given by
\(\frac{7 x+y+14}{\sqrt{7^2+1^2}}\) = \(\pm \frac{x+7 y-7}{\sqrt{1^2+7^2}}\)
⇒ 7x + y + 14 = ± (x + 7y – 7)
i.e. 6x – 6y + 21 = 0
⇒ 2x – 2y + 7 = 0 …(5)
and 7x + y + 14 = – x – 7y + 7
⇒ 8x + 8y + 7 = 0 …(6)
On solving the eqns. of AB and BC be given by y = \(\frac { 1 }{ 3 }\) and x = \(\frac { 14 }{ 3 }\)
∴ Coordinates of B are \(\left(\frac{14}{3}, \frac{1}{3}\right) \text {. }\)
For point A \(\left(\frac{-19}{6}, \frac{49}{6}\right)\);
2 × \(\left(\frac{-19}{6}\right)\) – 2\(\left(\frac{49}{6}\right)\) + 7 = \(\frac{-38-98+42}{6}\) < 0 For point B\(\left(\frac{14}{3}, \frac{1}{3}\right)\); 2\(\left(\frac{14}{3}\right)\) – \(\frac{2}{3}\) + 7 > 0
Thus points A and B lies on opposite sides of eqn. (5).
∴ eqn. (5) will become the internal bisector of angle C. Thus eqn. (6) gives the bisector of exterior angle at C.

Question 8.
Find the centre of the inscribed circle of the triangle the equations of whose sides are y – 15 = 0, 12y + 5x = 0 and 4y – 3x = 0.
Solution:
Given eqns. of sides of triangle are
y – 15 = 0 …(1)
5x + 12y = 0 …(2)
and 4y – 3x = 0 …(3)
lines (1) and (2) intersects at C(- 36, 15)
lines (2) and (3) intersects at A(0, 0)
and lines (1) and (3) intersects at B (20, 15)

The eqns. of bisectors of the angle opposite to y – 15 = 0 are given by
\(\frac{4 y-3 x}{\sqrt{4^2+(-3)^2}}\) = \(\pm \frac{5 x+12 y}{\sqrt{5^2+12^2}}\)
⇒ \(\frac{4 y-3 x}{5}\) = \(\pm \frac{5 x+12 y}{13}\)
⇒ 52y – 39x = 25x + 60y
⇒ 64x + 8y = 0
⇒ 8x + y = 0 …(4)
and 52y – 39x = – 25x – 80y
⇒ 14x – 112y = 0
⇒ x – 8y = 0 …(5)
For point B(20, 15);
8 × 20 + 15 = 175 > 0
For point C(- 36, 15);
8 × (- 36) + 15 < 0
Thus the points B and C are lies on opposite sides of eqn. (4) and hence eqn. (4) will be the internal bisector of angle A. The internal bisectors of angle B are given by
\(\frac{4 y-3 x}{\sqrt{4^2+(-3)^2}}\) = \(\pm \frac{y-15}{1}\)
⇒ \(\frac{4 y-3 x}{5}\) = ± y – 15
i.e. 3x + y – 75 = 0 …(6)
and 3x – 9y + 75 = 0
⇒ x – 3y + 25 = 0 …(7)
For point A(0, 0);
L.H.S of eqn. (7) = 0 + 25 = 25 > 0
For point C(- 36, 15);
L.H.S = – 36 – 45 + 25 = – 56 < 0
Thus points A and C are lies on opposite sides of eqn. (7) and hence will be the internal bisector of angle B.
The internal bisector of angle C are given by
\(\frac{5 x+12 y}{\sqrt{5^2+12^2}}\) = ± (y – 15)
⇒ 5x – y + 195 = 0 …(8)
and 5x + 25y – 195 = 0
⇒ x + 5y – 39 = 0 …(9)
For point A(0, 0);
L.H.S of eqn. (9) = 0 + 0 – 39 = – 39 < 0 For point B (20, 15); L.H.S = 20 + 75 – 39 = 56 > 0
Thus points A and B are lies on opposite sides of eqn. (9) and it will be the internal bisector of angle C.
Clearly the centre of the inscribed circle of the triangle will be the Incentre i.e. the point of concurrence of all three internal bisectors of angles of Δ.
On solving eqn. (4) and (7) ; we have
x = – 1 and y = 8
The point (- 1, 8) also lies on eqn. (9).
Hence all the three internal bisectors intersects at point (- 1, 8) and will be the required centre of inscribed circle.

Question 9.
The co-ordinates of A, B, C are respectively (- 4, 0),(0, 2) and (- 3, 2). Find (i) the equation of the straight line which bisects the angle CAB internally; (ii) the co-ordinates of the point where this straight line meets the straight line joining C to the middle point of AB.
Solution:
eqn. of line AB be given by
y – 0 = \(\frac{2-0}{0+4}\) (x + 4)
⇒ y = \(\frac { 1 }{ 2 }\) (x + 4)
⇒ x – 2y + 4 = 0 …(1)

and eqn. of line AC be given by
y – 0 = \(\frac{2-0}{-3+4}\) (x + 4)
⇒ y = 2x + 8 ⇒ 2x – y + 8 = 0 ..(2)
The eqns. of bisectors of angle A be given by
\(\frac{x-2 y+4}{\sqrt{1^2+(-2)^2}}\) = \(\pm \frac{2 x-y+8}{\sqrt{2^2+(-1)^2}}\)
Taking +ve sign ;
x – 2y + 4 = 2x – y + 8
⇒ x + y + 4 = 0 …(3)
Taking -ve sign ;
x – 2y + 4 = – 2x + y – 8
⇒ 3x – 3y + 12 = 0
⇒ x – y + 4 = 0 …(4)
putting the coordinates of B and C in L.H.S of eqn. (4); we have
For point B(0, 2) ; 0 – 2 + 4 = 2 > 0
For point C(- 3, 2); – 3 – 2 + 4 = – 1 < 0
Thus points B(0, 2) and C(- 3, 2) are lies on opposite sides of eqn. (4).
Thus eqn. (4) will be the eqn. of internal bisector of angle A i.e. angle CAB.

(ii) Middle point of AB be given by
\(\mathrm{D}\left(\frac{-4+0}{2}, \frac{0+2}{2}\right)\) i.e. D(-2, 1)
Thus using two point form, eqn. of line CD be given by
y – 2 = \(\frac{1-2}{-2+3}\)(x + 3)
⇒ y – 2 = -(x + 3)
⇒ x + y + 1 = 0 …(5)
On solving eqn. (4) and eqn. (5); we have
x = –\(\frac { 5 }{ 2 }\) and y = \(\frac { 3 }{ 2 }\)
Thus required point of intersection be \(\left(-\frac{5}{2}, \frac{3}{2}\right)\).